CE 394K.2 Hydrology

Homework Problem Set #3

Due Thurs March 29

Problems in "Applied Hydrology"

4.2.3 Infiltration by Horton's method

4.3.2 Infiltration by Green-Ampt method

4.4.3 Ponding time and cumulative infiltration at ponding

4.4.11 Theoretical study of infiltration at ponding time using Philip's equation

( 1 ) (2) (3)Inr11 tration

Timet

(hr)

Rate Cumulative. f F

(in/hr) (1n)

0.00.51.01.52.0

0.000.781. 111.381.65

3.000.8J10.570.530.53

Table JI.2.1. Infiltration computed by Horton's equation

".2.2.

Assuming continuously ponded conditions, the cumulative infiltration attime t · 0.75 hrs for Horton's equation is g1ven by Eq. (2) from Table JI.JI.1of the textbook

F . r t + (f o - f )(1 - e-kt)/kc c. so that

F(0.75) · 0.53 x 0.75 + (3 - 0.53)[1 - exp(-1I.182x 0.75)]/11.182

· 0.96 in.

The cumulative infUtration at time t · 2 hr can be similarly computed,giving F(2) · 1.65 in. Therefore, the incremental depth of infiltrationbetween time t · 0.75 and t · 2 hrs. is 1.65 - 0.96 · 0.69 in.

suming continuously ponded conditions, the values of the infiltrationrate r and the cumulative infiltration F are computed using Eqs. (1) and (~)from Table 11.11.1of the textbook,

f(t)

F(t)

· fc + (fo - fc) e-kt

· fct + (fo - fc)(1 - e-kt)/kwith fc - 1 cm/hr, fo · 5 cm/hr and k. 2 hr-1.

4-8

(a) Infiltration vs. Time.

o , 2 I .InflltNllen D8pIh (...)

Fig. 4.2.3. Infiltrationrate and cumulative infiltration depth computed byHorton's equation.

4-10

I.

! I

1.i 2-I ,

0 .0 0.2 0.. 0.' u , 1.2 ''''' I.' ,.I 2

1'bn8eM)

(b) Rate vs. Infiltration Depth.

....

..

11.1

i I

I

I2.1

2

,.I

,

(1) (2) (3)Infll tration

Timet

(hr)

Rate Cumulativef F

(cm/hr) (em)

0.00.51.01.52.0

s.oo2.1J71. 51J1.191.07

0.001. 762.723.1JO3.96

Table 11.2.3. Infiltrationcomputed by Horton's equation.

The resul ts are shown 1n Table Ji.2.3 for time t · 0, O.S, 1, 1.5 and 2hrs. For example, for time t · O.S hrs, f(O.S). 1 + (S - 1) exp(-2 x O.S)· 0.81f in/hr, as shown in Col. (2) ot Table 11.2.3and F(O.S) · 1 x O.S + (S- 1)[ 1 - exp(-2 x O.S)1/2 · 0.78 in, as shown in Col. (3) of 'the table.

The infiltration rate and cumulative infiltration rate are plottedversus time in Fig. 1I.2.3(a). Fig. 1I.2.3(b) shows the InfU tration rate asa function of cumulatlve infiltration.

1J.2.1J.

Assuming continuously ponded conditions, the infiltration rate is,according to Horton's equation (Eq. 11.2.3 from the textbook)

f · fc + (fo - fc) e-kt

so that

fe · (f - fo e-kt)/(1 - e-kt)

The cumulative infiltration for Horton's equation is given by

F · ret + (fo - fe)(1 - e-kt)/kSubstituting fo in the previous equation yields

4-11

- --- -

@)The infiltration rate f and the cumulative infiltration F at time t ·

O. 0.5. 1, 1.5, 2, 2.5 and 3 hrs may be computed following the method-'outlined In Problem 4.3.1. The cumulative infiltration is first computedusing Eq. (4.3.8) of the textbook

F(t) · kt + .A8 1n[1 + F/(.A8)]

· 1.09 t + 2.72 1n[1 + 3/2.72]

which may be solved by successive approximation for each value of t. TheinfUtration rate is then computed using Eq. (4.3.7) of the textbook

f(t) · k (.A8/F + 1) · 1.09 (2.72/F + 1)

The results are listed in Table 4.3.2. Fig. 4.3.2(a) shows a plot of theinf 11 tration rate and cumulati ve inf 11 tration versus ti me. Fig. 4.3.2(b)

shows the variation of the infiltration rate f with the infiltration depthF.

e .. ().4\2..

4' ~. (t.o.k ';) f.~'

A~ .. ~.. (l- ~«.)

= 0--4t).(l~ .4)'=' t).}.A.1 ').

""A~ ~ 'Cot 'C~.'Uc'.l

:: ~-1'2

:D...t.~ --r~

~ '\. ( p. t\~ ~ +»tk

InfUtration

Timet(hr)

Rater

( cm/hr )

DepthF(em)

Table 4.3.2. Infiltration

computedby theGreen-Amptmethod.

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0.0 . 0.000.5 2.51 2.101.0 2.01 3.211.5 1.80 11.162.0 1.68 5.032.5 1.60 5.85

_ 3.0 1.54 6.63

(a) Infiltration vs. Time

.....

0.. t.l t..

c )

(b) Rate vs. Infiltration Depth

2 2.1

W'DI. D8pIh (em)

Fig. 4.3.2. Infiltration rate andeumulative infiltration depth eomputedby the Green-Ampt method.

4-28

- ------

to

.

f.

7""'

j .

1 ..

!4

I I

2

t

00

.

t2

tt

to

I.1""'

! 7

I

.

.

.

.2

10 :& . .

"'

exam,;l e, starting with F - 1 em gi ves a new val ue F - 0.97 + 6.~ 8R.n[(6.!!8+1)/(6.~8+0.97») + 0.65(1 - 0.19) - 1.52 em. This value is thensubstituted in the right hand side of the previous equation and a new valueF · 1.96'em 1s obtained. After 17 iterations, the solution converges to F ·3.17 em. The eorresponding infUtration rate is given by Eq. ~.3.t from thetextbook .

f · K(1 + ~~e/F) · 0.65 (1 + 6.48/3.17) - 1.98 cm/hr

~.

~ a clay loam soil, from Table 11.3.1 of the textbook, ee · 0.309, ~ -20.88 em and K · 0.1 em/hr. The ini tial.effeeti ve saturation is S~ · 0.25so from Eq. .(11.3.10) from the textbook, ~e · (1 - Se)ee · (1 - 0.25)0.309.0.232 and ~l1e · 20.88 x 0.232 · 4.8~ em. For i · 1 em/hr, the ponding timeis gi ven by Eq. (Jf.Jf.2) of the textbook

tp - K~~e/[i(1-K)] - .0.1 x lI.8Jf/[1 (1 - 0.1)] - 0.5l1 hr

and the infUtrated depth at ponding is Fp · tpi · 0.511 x'1 - 0.5~ em.

For i · 3 em/hr, tp and Fp may be similarly eomputed to y1eld.1p. - 0.06J1J:.and Fp.O. 17 em._

4.II.Jf.

From Problem Jf.Jf.3, K · 0.1 em/hr, ~l1e · Jf.8!! em, tp · 0.06 hr and FR-0~17 em under rainfall intensity i - 3 em/hr. For t · 1 hr, theinfiltration depth 1s given by Eq. (4.11.5) from the textbook

F · Fp + ~~e In[(~l1e+F)/(~~e+Fp)] + K(t - tp)

· 0.17 + 1I.8!! R.n[(lI.8l1+F)/(4.8l1+0.17)] + 0.1(1 - 0.06)

The solution F may be found by the method of sueeessive substitution. Forexample, starting with F · 1 em gives a new value F · 0.11 + lI.8l!In[(lI.8!1+1)/(1I.8l1+0.17)] + 0.1(1 - 0.06) · 1.01 em. This value is thensubstituted in the right hand side of the previous equation and a new valueof F is obtained. The solution eonverges to F . 1.0Jf em. The eorrespondinginfiltration rate is given by Eq. (lI.3.7) from the textbook

.f. K(1 + ~~e/F) · 0.1 (1 + lI.8l1/1.0ll) · 0..57 em/hr

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tp . S2(I-K/2)/[21(I-K)2J . 52 (6 - 0.4/2)/[2 x 6 (6-0.~)2J· 0.385 hr

The eumulative infiltration at ponding is Fp · itp · 6 x 0.385 · 2.31 em.

4.11.10.

The ponding time for the Horton's equation is given in Table ~.4.1 of

the textbook. For fo · 10 em/hr, fe · ~ em/hr, k · 2 hr-1 and rainfall ~intensityi · 6 em/hr, thIs gives ~

tp · {fo-!+fe 1n«fo - fe)/(1 - fo)]}/(1k>~

· (10 - 6 + 11 In[(10-1I)/(6-II)J}/(6 x 2) · 0.70 hr

The eumulative infiltration at pending is Fp · Itp · 6 x 0.70 · 4.2 em.

For Philip's equation (Table 11.11.1of the textbook), the infiltrationrate is

f . (S/2) t-1/2 + K

so that the time t may be expressed as

Substituting t into the equation for the cumulative inf 11tration yields

F . St1/2 + Kt . SI/[2(f-K)] + KS1/[4(f-K)IJ - SI(f-K/2)/[2(f-K)2J

The cumulative infiltration at pondlng time is Fp · itp and the infIltrationrat~ is t · i, where i is the constant rainfall rate (see Fig. 4.4.2 trom

the textbgok). Substitut1ng Fp and fp into the prev10us equation yields

itp . SI(1-K/2)/[2(1-K)IJ

so that

4.4.12.

The infiltrat10n rate for Horton's equation is, from Table 4.11.1of thetextbook,

f · fa + (fo-fo> e-Kt

so that the time t maybe expreaaed aa

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